integral from 1 to 8 of (\sqrt[3]{x}+1)

解答

\int_{1}^{8} (3 x + 1) d x

\frac{73}{4}

十进制

18.25

求解步骤

\int_{1}^{8} 3 x + 1 d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{8} 3 x d x + \int_{1}^{8} 1 d x

\int_{1}^{8} 3 x d x = \frac{45}{4}

\int_{1}^{8} 1 d x = 7

= \frac{45}{4} + 7

化简

\frac{45}{4} + 7 : \frac{73}{4}

= \frac{73}{4}

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